The work done by `1` mole of ideal gas during an adiabatic process is (are ) given by :

Options

A. `(P_(2)V_(2) - P_(1)V_(1))/(gamma-1)`
B. `(nR(T_(1)-T_(2)))/(gamma-1)`
C. `(P_(2)V_(2)-P_(1)V_(1))/(gamma)`
D. None of these

Explanation

To find the work done by 1 mole of an ideal gas during an adiabatic process, we can follow these steps:

### Step 1: Understand the Adiabatic Process
An adiabatic process is one in which there is no heat exchange with the surroundings. This means that the heat transfer \( Q = 0 \).

### Step 2: Use the Formula for Work Done
The work done \( W \) by an ideal gas during an adiabatic process can be expressed as:
\[ W = \frac{nR}{\gamma - 1} (T_f - T_i) \]
where:
- \( n \) is the number of moles of gas (which is 1 mole in this case),
- \( R \) is the universal gas constant,
- \( \gamma \) is the ratio of specific heats \( \frac{C_p}{C_v} \),
- \( T_f \) is the final temperature,
- \( T_i \) is the initial temperature.

### Step 3: Substitute Values
Since we are given that \( n = 1 \), we can simplify the equation:
\[ W = \frac{R}{\gamma - 1} (T_f - T_i) \]
or equivalently,
\[ W = \frac{R}{\gamma - 1} (T_2 - T_1) \]
where \( T_2 \) is the final temperature and \( T_1 \) is the initial temperature.

### Step 4: Relate to Ideal Gas Law
Using the ideal gas law \( PV = nRT \), we can express the work done in terms of pressure and volume:
\[ W = \frac{P_2V_2 - P_1V_1}{\gamma - 1} \]
This is derived from the fact that \( nRT_1 = P_1V_1 \) and \( nRT_2 = P_2V_2 \).

### Step 5: Final Expression
Thus, the work done by 1 mole of an ideal gas during an adiabatic process can be expressed as:
\[ W = \frac{P_2V_2 - P_1V_1}{\gamma - 1} \]

### Conclusion
The correct answer to the question is:
\[ W = \frac{P_2V_2 - P_1V_1}{\gamma - 1} \]